9. 使分式$\frac{x - 2}{x^{2}-4x + 4}=\frac{1}{2 - x}$自左向右变形成立的条件是(
A.$x ＞ -2$
B.$x ＜ 2$
C.$x ≥ 2$
D.$x ≤ 2$
B
)A.$x ＞ -2$
B.$x ＜ 2$
C.$x ≥ 2$
D.$x ≤ 2$
答案：9. B 解析:$\because \frac {|x-2|}{x^{2}-4x+4}=\frac {|x-2|}{(x-2)^{2}}=\frac {1}{2-x},\therefore x-2＜0,\therefore x＜2$.故选B.
10. 已知分式$\frac{2x}{3x^{2}+5y^{2}}$的值为 9.
(1)若把分式中的$x,y$同时扩大为原来的 3 倍,则分式的值为
(2)若把分式中的$x,y$同时缩小为原来的$\frac{1}{n}$后,分式的值是 27,则$n=$
(1)若把分式中的$x,y$同时扩大为原来的 3 倍,则分式的值为
3
;(2)若把分式中的$x,y$同时缩小为原来的$\frac{1}{n}$后,分式的值是 27,则$n=$
3
.答案：10. (1)3 解析:当分式中的x,y同时扩大为原来的3倍时,$\frac {2x}{3x^{2}+5y^{2}}=\frac {3×2x}{9(3x^{2}+5y^{2})}=\frac {1}{3}×\frac {2x}{3x^{2}+5y^{2}}=3$.
(2)3 解析:$\because \frac {2x}{3x^{2}+5y^{2}}=9,\therefore \frac {\frac {2x}{n}}{3(\frac {x}{n})^{2}+5(\frac {y}{n})^{2}}=\frac {2nx}{3x^{2}+5y^{2}}=n· \frac {2x}{3x^{2}+5y^{2}}=9n=27,\therefore n=3$.
(2)3 解析:$\because \frac {2x}{3x^{2}+5y^{2}}=9,\therefore \frac {\frac {2x}{n}}{3(\frac {x}{n})^{2}+5(\frac {y}{n})^{2}}=\frac {2nx}{3x^{2}+5y^{2}}=n· \frac {2x}{3x^{2}+5y^{2}}=9n=27,\therefore n=3$.
11. 已知$\frac{1}{x}+\frac{1}{y}=3$,求分式$\frac{2x - 3xy + 2y}{x + 2xy + y}$的值.
答案：11. 分式的分子、分母都除以xy,则$\frac {2x-3xy+2y}{x+2xy+y}=\frac {\frac {2x-3xy+2y}{xy}}{\frac {x+2xy+y}{xy}}=$$\frac {\frac {2}{y}-3+\frac {2}{x}}{\frac {1}{y}+2+\frac {1}{x}}.\because \frac {1}{x}+\frac {1}{y}=3$,
∴ 原式$=\frac {2×3-3}{3+2}=\frac {3}{5}$.
∴ 原式$=\frac {2×3-3}{3+2}=\frac {3}{5}$.
12. 若$x^{2}-3x - 1 = 0$,求下列各式的值.
(1)$\frac{x^{2}-1}{x}$;
(2)$\frac{x^{4}+1}{x^{2}}$;
(3)$\frac{2x^{2}}{x^{4}-3x^{2}+1}$.
(1)$\frac{x^{2}-1}{x}$;
(2)$\frac{x^{4}+1}{x^{2}}$;
(3)$\frac{2x^{2}}{x^{4}-3x^{2}+1}$.
答案：12. (1)解法一:$\frac {x^{2}-1}{x}=x-\frac {1}{x},\because x^{2}-3x-1=0$,等式两边同除以x,即$x-3-\frac {1}{x}=0,\therefore x-\frac {1}{x}=3$,即$\frac {x^{2}-1}{x}=3$.
解法二:$\because x^{2}-3x-1=0,\therefore x^{2}-1=3x$,
∴ 原式$=\frac {3x}{x}=3$.
(2)$\frac {x^{4}+1}{x^{2}}=x^{2}+\frac {1}{x^{2}}=x^{2}-2+\frac {1}{x^{2}}+2=(x-\frac {1}{x})^{2}+2=3^{2}+2=11$.
(3)分式的分子、分母都除以$x^{2}$,则原式$=\frac {2}{x^{2}-3+\frac {1}{x^{2}}}=$$\frac {2}{11-3}=\frac {1}{4}$.
解法二:$\because x^{2}-3x-1=0,\therefore x^{2}-1=3x$,
∴ 原式$=\frac {3x}{x}=3$.
(2)$\frac {x^{4}+1}{x^{2}}=x^{2}+\frac {1}{x^{2}}=x^{2}-2+\frac {1}{x^{2}}+2=(x-\frac {1}{x})^{2}+2=3^{2}+2=11$.
(3)分式的分子、分母都除以$x^{2}$,则原式$=\frac {2}{x^{2}-3+\frac {1}{x^{2}}}=$$\frac {2}{11-3}=\frac {1}{4}$.
13. 阅读材料题:
已知$\frac{a}{3}=\frac{b}{4}=\frac{c}{5}$,求分式$\frac{2a + 3b - c}{a - b + 2c}$的值.
解:设$\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=k$,则$a = 3k,b = 4k,c = 5k$,
所以$\frac{2a + 3b - c}{a - b + 2c}=\frac{6k + 12k - 5k}{3k - 4k + 10k}=\frac{13k}{9k}=\frac{13}{9}$.
参照上述材料解题:
(1)已知$\frac{x}{2}=\frac{y}{3}=\frac{z}{6}$,求分式$\frac{x + 2y - z}{x - 2y + 3z}$的值;
(2)已知$\frac{y + z}{x}=\frac{z + x}{y}=\frac{x + y}{z}$,其中$x + y + z ≠ 0$,求$\frac{x + y - z}{x + y + z}$的值.
已知$\frac{a}{3}=\frac{b}{4}=\frac{c}{5}$,求分式$\frac{2a + 3b - c}{a - b + 2c}$的值.
解:设$\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=k$,则$a = 3k,b = 4k,c = 5k$,
所以$\frac{2a + 3b - c}{a - b + 2c}=\frac{6k + 12k - 5k}{3k - 4k + 10k}=\frac{13k}{9k}=\frac{13}{9}$.
参照上述材料解题:
(1)已知$\frac{x}{2}=\frac{y}{3}=\frac{z}{6}$,求分式$\frac{x + 2y - z}{x - 2y + 3z}$的值;
(2)已知$\frac{y + z}{x}=\frac{z + x}{y}=\frac{x + y}{z}$,其中$x + y + z ≠ 0$,求$\frac{x + y - z}{x + y + z}$的值.
答案：13. (1)设$\frac {x}{2}=\frac {y}{3}=\frac {z}{6}=k$,则$x=2k,y=3k,z=6k,\therefore \frac {x+2y-z}{x-2y+3z}=$$\frac {2k+6k-6k}{2k-6k+18k}=\frac {2k}{14k}=\frac {1}{7}$,
∴ 分式$\frac {x+2y-z}{x-2y+3z}$的值为$\frac {1}{7}$.
(2)设$\frac {y+z}{x}=\frac {z+x}{y}=\frac {x+y}{z}=k$,则$\{\begin{array}{l} y+z=kx,①\\ x+z=ky,②\\ x+y=kz,③\end{array} $①+②+③得$2x+2y+2z=k(x+y+z).\because x+y+z≠0,\therefore k=2$,
∴ 原式=$\frac {2z-z}{2z+z}=\frac {z}{3z}=\frac {1}{3}$.
∴ 分式$\frac {x+2y-z}{x-2y+3z}$的值为$\frac {1}{7}$.
(2)设$\frac {y+z}{x}=\frac {z+x}{y}=\frac {x+y}{z}=k$,则$\{\begin{array}{l} y+z=kx,①\\ x+z=ky,②\\ x+y=kz,③\end{array} $①+②+③得$2x+2y+2z=k(x+y+z).\because x+y+z≠0,\therefore k=2$,
∴ 原式=$\frac {2z-z}{2z+z}=\frac {z}{3z}=\frac {1}{3}$.
14. 已知三个数$x,y,z$满足$\frac{xy}{x + y}=-3,\frac{yz}{y + z}=\frac{4}{3},\frac{zx}{z + x}=-\frac{4}{3}$,求$\frac{xyz}{xy + yz + zx}$的值.
答案：14. $\because \frac {xy}{x+y}=-3,\frac {yz}{y+z}=\frac {4}{3},\frac {zx}{z+x}=-\frac {4}{3},\therefore \frac {x+y}{xy}=-\frac {1}{3},\frac {y+z}{yz}=\frac {3}{4},$$\frac {z+x}{zx}=-\frac {3}{4}$,即$\frac {1}{x}+\frac {1}{y}=-\frac {1}{3},\frac {1}{y}+\frac {1}{z}=\frac {3}{4},\frac {1}{z}+\frac {1}{x}=-\frac {3}{4}$,解得$\frac {1}{x}+\frac {1}{y}+\frac {1}{z}=-\frac {1}{6},\therefore \frac {xyz}{xy+yz+zx}=\frac {1}{\frac {1}{z}+\frac {1}{x}+\frac {1}{y}}=-6$.
解析：
$\because \frac{xy}{x + y}=-3,\frac{yz}{y + z}=\frac{4}{3},\frac{zx}{z + x}=-\frac{4}{3}$，
$\therefore \frac{x + y}{xy}=-\frac{1}{3},\frac{y + z}{yz}=\frac{3}{4},\frac{z + x}{zx}=-\frac{3}{4}$，
即$\frac{1}{x}+\frac{1}{y}=-\frac{1}{3}$，$\frac{1}{y}+\frac{1}{z}=\frac{3}{4}$，$\frac{1}{z}+\frac{1}{x}=-\frac{3}{4}$。
将上述三式相加得：$2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=-\frac{1}{3}+\frac{3}{4}-\frac{3}{4}=-\frac{1}{3}$，
$\therefore \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{1}{6}$。
$\because \frac{xy + yz + zx}{xyz}=\frac{1}{z}+\frac{1}{x}+\frac{1}{y}=-\frac{1}{6}$，
$\therefore \frac{xyz}{xy + yz + zx}=\frac{1}{-\frac{1}{6}}=-6$。
$\therefore \frac{x + y}{xy}=-\frac{1}{3},\frac{y + z}{yz}=\frac{3}{4},\frac{z + x}{zx}=-\frac{3}{4}$，
即$\frac{1}{x}+\frac{1}{y}=-\frac{1}{3}$，$\frac{1}{y}+\frac{1}{z}=\frac{3}{4}$，$\frac{1}{z}+\frac{1}{x}=-\frac{3}{4}$。
将上述三式相加得：$2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=-\frac{1}{3}+\frac{3}{4}-\frac{3}{4}=-\frac{1}{3}$，
$\therefore \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{1}{6}$。
$\because \frac{xy + yz + zx}{xyz}=\frac{1}{z}+\frac{1}{x}+\frac{1}{y}=-\frac{1}{6}$，
$\therefore \frac{xyz}{xy + yz + zx}=\frac{1}{-\frac{1}{6}}=-6$。